有关系时如何汇总数据集中的前 3 个最高值

Question

我有一个数据框 (my_data)，即使可能存在联系，也只想计算 3 个最高值的总和。我对 R 很陌生，我已经使用了dplyr.

A tibble: 15 x 3
   city      month number
   <chr>     <chr>  <dbl>
 1 Lund      jan       12
 2 Lund      feb       12
 3 Lund      mar       18
 4 Lund      apr       28
 5 Lund      may       28
 6 Stockholm jan       15
 7 Stockholm feb       15
 8 Stockholm mar       30
 9 Stockholm apr       30
10 Stockholm may       10
11 Uppsala   jan       22
12 Uppsala   feb       30
13 Uppsala   mar       40
14 Uppsala   apr       60
15 Uppsala   may       30

这是我试过的代码：

# For each city, count the top 3 of variable number
my_data %>% group_by(city) %>% top_n(3, number) %>% summarise(top_nr = sum(number))

预期的（想要的）输出是：

# A tibble: 3 x 2
  city      top_nr
  <chr>      <dbl>
1 Lund          86
2 Stockholm     75
3 Uppsala      130

但实际的 R 输出是：

# A tibble: 3 x 2
  city      top_nr
  <chr>      <dbl>
1 Lund          86
2 Stockholm     90
3 Uppsala      160

似乎如果有平局，所有平局值都包含在总和中。我只想计算具有最高值的 3 个唯一实例。

任何帮助将非常感激！:)

Answer 1

我们可以做 adistinct来删除重复的元素。在这种方式top_n的工作原理是，如果值是重复的，它会继续，许多受骗者行

my_data %>% 
   distinct(city, number, .keep_all = TRUE) %>%
   group_by(city) %>%
   top_n(3, number) %>%
   summarise(top_nr = sum(number))

更新

基于OP的新输出，在top_n输出（不是arranged）之后，得到按降序排列的'number'，并得到sum前3个'number'的

my_data %>% 
   group_by(city) %>% 
   top_n(3, number) %>% 
   arrange(city,  desc(number)) %>% 
   summarise(number = sum(head(number, 3)))
# A tibble: 3 x 2
#  city      number
#  <chr>      <int>
#1 Lund          74
#2 Stockholm     75
#3 Uppsala      130

数据

my_data <- structure(list(city = c("Lund", "Lund", "Lund", "Lund", "Lund", 
"Stockholm", "Stockholm", "Stockholm", "Stockholm", "Stockholm", 
"Uppsala", "Uppsala", "Uppsala", "Uppsala", "Uppsala"), month = c("jan", 
"feb", "mar", "apr", "may", "jan", "feb", "mar", "apr", "may", 
"jan", "feb", "mar", "apr", "may"), number = c(12L, 12L, 18L, 
28L, 28L, 15L, 15L, 30L, 30L, 10L, 22L, 30L, 40L, 60L, 30L)), 
class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15"))

Answer 2

如果没有top_n()：生活可能会更简单：

dat %>%
  group_by(city) %>%
  summarize(
    top_nr = sum(tail(sort(number), 3))
    )